Palaksha

Shakunthala, V.

Suchitra, G.

University of Mysore - Digital Repository of Research, Innovation and Scholarship (ePrints@UoM)

Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 33 (2013) no. 2, 35–42. doi:10.1285/i15900932v33n2p35On some continued fraction expansions forthe ratios of the function ρ(a, b)D. D. SomashekaraDepartment of Studies in Mathematics, University of Mysore, Manasagangotri,Mysore 570 006, India.dsomashekara@yahoo.comS. L. ShaliniDepartment of Studies in Mathematics, University of Mysore, Manasagangotri,Mysore 570 006, India.shalinisl.maths@gmail.comK. Narasimha MurthyDepartment of Studies in Mathematics, University of Mysore, Manasagangotri,Mysore 570 006, India.simhamurth@yahoo.comReceived: 7.1.2013; accepted: 1.4.2013.Abstract. In his lost notebook, Ramanujan has defined the function ρ(a, b) byρ(a, b) :=(1 +1b) ∞∑n=0(−1)nqn(n+1)/2anb−n(−aq)n,where |q| < 1, and (a; q)n =∏n−1k=0 (1 − aqk), n = 1, 2, 3, . . . , and has given a beautiful reci-procity theorem involving ρ(a, b). In this paper we obtain some continued fraction expansionsfor the ratios of ρ(a, b) with some of its contiguous functions. We also obtain some interest-ing special cases of our continued fraction expansions which are analogous to the continuedfraction identities stated by Ramanujan.Keywords: Basic hypergeometric series, q-continued fractions.MSC 2010 classification: 2010 Mathematics Subject Classification: primary 33D05,secondary 11A55.1 IntroductionRamanujan, a pioneer in the theory of continued fractions has recordedscores of continued fraction identities in chapter 12 of his second notebook[23] and in his lost notebook [24]. This part of Ramanujan’s work has beentreated and developed by several authors including Andrews [4], Hirschhorn [19],Carlitz [12], Gordon [18], Al-Salam and Ismail [3], Ramanathan [21], [22], Denishttp://siba-ese.unisalento.it/ c© 2013 Universita` del Salento36 D. D. Somashekara, S. L. Shalini and K. Narasimha Murthy[13], [14], [15], Bhargava and Adiga [8], [9], Bhargava, Adiga and Somashekara[10], [11], Adiga and Somashekara [2], Verma, Denis and Srinivasa Rao [29],Singh [26], Bhagirathi [5], [6], [7], Adiga, Denis and Vasuki [1], Denis, Singhand Bhagirathi [17], Denis and Singh [16], Vasuki [27], Vasuki and Madhusudan[28], Somashekara and Fathima [25], Mamta and Somashekara [20].The main purpose of this paper is to establish continued fraction expansionsfor the ratios ρ(aq, b)/ρ(a, b) and ρ(a, bq)/ρ(a, b), whereρ(a, b) =(1 +1b) ∞∑n=0(−1)nqn(n+1)/2anb−n(−aq)n , (1.1)which was given by Ramanujan in his lost notebook [24]. In fact Ramanujanhas given a beautiful reciprocity theorem for the function ρ(a, b) in his lostnotebook.In section 2, we prove some key functional relations satisfied by ρ(a, b), whichwill be used in the development of continued fractions. In section 3, we proveour main results and in section 4 we obtain some special cases of our continuedfractions which are analogous to the continued fractions of Ramanujan.2 Some functional relations satisfied by ρ(a, b)In this section, we prove some functional relations satisfied by ρ(a, b).Lemma 1. ρ(a, b) satisfies the following functional relations.(1 + aq)ρ(a, b)(1 + 1b) − aqρ(aq, b)(1 + 1b) = ρ(aq, bq)(1 + 1bq) , (2.1)(1 + aq)ρ(a, bq)(1 + 1bq) − (1 + aq) ρ(a, b)(1 + 1b) = aqbρ(aq, b)(1 + 1b) − abρ(aq, bq)(1 + 1bq) , (2.2)ρ(a, bq)(1 + 1bq) = (1− ab) ρ(a, b)(1 + 1b) + aqb (1 + a)(1 + aq)ρ(aq, b)(1 + 1b) , (2.3)ρ(a, b) =(1− aqb + aq1 + aq)ρ(aq, b) +(aq2/b1 + aq2)ρ(aq2, b), (2.4)(1 + aq)ρ(a, bq)(1 + 1bq) − aqρ(aq, bq)(1 + 1bq) = ρ(aq, bq2)(1 + 1bq2) , (2.5)ρ(a, b) =[a+ bq(a− 1)a(1 + bq)]ρ(a, bq) +[bq2a(1 + bq2)]ρ(a, bq2) (2.6)On some continued fraction expansions for the ratios of the function ρ(a, b) 37Proof. Using (1.1), the left side of (2.1) can be written as(1 + aq) + (1 + aq)∞∑n=1(−1)nqn(n+1)/2anb−n(−aq)n− aq − aq∞∑n=1(−1)nqn(n+1)/2(aq)nb−n(−aq2)n= 1 +∞∑n=1(−1)nqn(n+1)/2anb−n(−aq2)n−1{1− aqn+11 + aqn+1}=ρ(aq, bq)(1 + 1bq) ,which is the right side of (2.1).Using (1.1), the left side of (2.2) can be written as(1 + aq)∞∑n=1(−1)nqn(n+1)/2an(bq)−n(−aq)n − (1 + aq)∞∑n=1(−1)nqn(n+1)/2anb−n(−aq)n=−ab∞∑n=1(−1)n−1qn(n−1)/2an−1b−n+1(−aq2)n−1+aqb∞∑n=1(−1)n−1qn(n−1)/2(aq)n−1b−n+1(−aq2)n−1=aqbρ(aq, b)(1 + 1bq) − abρ(aq, bq)(1 + 1bq) .This proves (2.2).Substituting for ρ(aq, bq)/(1 + 1/bq) in (2.2) from (2.1), we obtain (2.3) onsome simplifications.Changing a to aq in (2.3), then adding resulting equation to (2.1),we obtain (2.4).Using (1.1), the left side of (2.5) can be written as(1 + aq) + (1 + aq)∞∑n=1(−1)nqn(n+1)/2an(bq)−n(−aq)n− aq − aq∞∑n=1(−1)nqn(n+1)/2(aq)n(bq)−n(−aq2)n= 1 +∞∑n=1(−1)nqn(n+1)/2an(bq)−n(−aq2)n−1{1− aqn+11 + aqn+1}=ρ(aq, bq2)(1 + 1bq2) .which is the right side of (2.5).38 D. D. Somashekara, S. L. Shalini and K. Narasimha MurthyAdding (2.1), (2.2) and the negative of (2.5), we obtain (2.6) on somesimplifications. QED3 Main resultsIn this section, we deduce the continued fraction expansions for the ratiosρ(aq, b)/ρ(a, b) and ρ(a, bq)/ρ(a, b).Theorem 1. We haveρ(aq, b)ρ(a, b)=(1 + aq)N1+M1N2+M2N3 + · · ·MnNn+1 · · · , (3.1)whereMn =aqn+1b(1 + aqn),andNn =(1− aqnb+ aqn), n = 0, 1, 2, . . . .Proof. Changing a to aqn in (2.4), we obtainρ(aqn, b) =(1− aqn+1b + aqn+11 + aqn+1)ρ(aqn+1, b) +(aqn+2/b1 + aqn+2)ρ(aqn+2, b).This can be written asTn ≡ ρ(aqn, b)ρ(aqn+1, b)=(1− aqn+1b + aqn+11 + aqn+1)+(aqn+2/b1+aqn+2)Tn+1. (3.2)Iterating (3.2) with n = 0, 1, 2, . . . , and then taking reciprocals, we obtain (3.1)after some simplifications. QEDTheorem 2. We haveρ(a, bq)ρ(a, b)=(1− ab)(1 + bq)q(1 + b)+(1 + bq)M0q(1 + b)N1+q(1 + b)M1N2+M2N3 + · · ·MnNn+1 · · · ,(3.3)where Mn and Nn are as in theorem (3.1).Proof. Equation (2.3) can be written asρ(a, bq)ρ(a, b)=(1− ab)(1 + bq)q(1 + b)+aqb (1 + a)(1 + bq)q(1 + b)(1 + aq) ρ(a, b)ρ(aq, b). (3.4)Iterating (3.2) with n = 0, 1, 2, . . . , and substituting the resulting identity in(3.4), we obtain (3.3) after some simplifications. QEDOn some continued fraction expansions for the ratios of the function ρ(a, b) 39Theorem 3. We haveρ(a, bq)ρ(a, b)=a(1 + bq)A0+B0A1+B1A2 + · · ·BnAn+1 · · · , (3.5)whereAn = [a+ bqn+1(a− 1)],andBn =[abqn+2(1 + bqn+1)], n = 0, 1, 2, . . . .Proof. Changing b to bqn in (2.6), we obtain on some simplificationsρ(a, bqn) =[a+ bqn+1(a− 1)a(1 + bqn+1)]ρ(a, bqn+1) +[bqn+2a(1 + bqn+2)]ρ(a, bqn+2).This can be written asFn ≡ ρ(a, bqn)ρ(a, bqn+1)=[a+ bqn+1(a− 1)a(1 + bqn+1)]+[bqn+2a(1+bqn+2)]Fn+1. (3.6)Iterating (3.6) with n = 0, 1, 2, . . . , and then taking reciprocals, we obtain (3.5)after some simplifications. QED4 Some special casesIn this section, we derive the following special cases of (3.1), (3.3) and (3.5).∑∞n=0(−1)nqn(n+3)/2(−q2)n∑∞n=0(−1)nqn(n+1)/2(−q)n=1 + q1+q2(1 + q)1+q3(1 + q2)1 + · · · , (4.1)∑∞n=0qn(n+3)/2(q2)n∑∞n=0qn(n+1)/2(q)n=1− q1−q2(1− q)1−q3(1− q2)1− · · · , (4.2)∑∞n=0qn(n+1)/2(q2)n∑∞n=0qn(n−1)/2(q)n=1− q(2− q)−q(1− q)(1 + q − q2)− · · · , (4.3)∑∞n=0(−1)nqn(n+5)/2(−q3)n∑∞n=0(−1)nqn(n+3)/2(−q2)n=1 + q21+q3(1 + q2)1+q4(1 + q3)1 + · · · , (4.4)40 D. D. Somashekara, S. L. Shalini and K. Narasimha Murthy∑∞n=0qn(n−1)/2(q2)n∑∞n=0qn(n+1)/2(q2)n= 2− q(1− q)(1 + q − q2)−q2(1− q2)(1 + q2 − q3)− · · · , (4.5)∑∞n=0qn(n+1)/2(q2)n∑∞n=0qn(n+3)/2(q2)n= (1 + q)− q2(1− q)1−q3(1− q2)1− · · · , (4.6)∑∞n=0(−1)nqn(n−3)/2(−q2)n∑∞n=0(−1)nqn(n−1)/2(−q2)n=q − 1q+(1 + q)q2 + · · · , (4.7)∑∞n=0(−1)nqn(n+1)/2(−q3)n∑∞n=0(−1)nqn(n+3)/2(−q3)n= (1− q) + q2(1 + q2)(1− q2 + q3) + · · · , (4.8)∑∞n=0(−1)nqn(n−1)/2(−q)n∑∞n=0(−1)nqn(n+1)/2(−q)n=2q1+q2(1 + q)1+q3(1 + q2)1 + · · · , (4.9)∑∞n=0qn(n−1)/2(q)n∑∞n=0qn(n+1)/2(q)n=2q(1 + 2q)−q2(1 + q)(1 + 2q2)− · · · , (4.10)∑∞n=0(−1)nqn(n+1)/2(−q2)n∑∞n=0(−1)nqn(n+3)/2(−q2)n=21+(1 + q)(1− q + q2)+q2(1 + q2)(1− q2 + q3) + · · · , (4.11)∑∞n=0(−1)nqn(n−3)/2(−q)n∑∞n=0(−1)nqn(n−1)/2(−q)n=q(1 + q)1+q3(1 + q2)1+q4(1 + q3)1+· · · . (4.12)Proof. Setting a = 1 = b in (3.1) and using the definition (1.1) of ρ(a, b) weobtain (4.1) after some simplifications. Similarly we obtain (4.2), (4.3) and (4.4)from (3.1) for a = −1, b = 1; a = −1, b = q and a = q, b = 1 respectively.Setting a = −q, b = q in (3.3) and using the definition (1.1) of ρ(a, b) weobtain (4.5) after some simplifications. Similarly we obtain (4.6), (4.7) and (4.8)from (3.3) for a = −q, b = 1; a = q, b = q2 and a = q2, b = q respectively.Setting a = 1 = b in (3.5) and using the definition (1.1) of ρ(a, b) we obtain(4.9) after some simplifications. Similarly we obtain (4.10), (4.11) and (4.12)from (3.5) for a = −1, b = 1; a = q, b = 1 and a = 1, b = q respectively. QEDAcknowledgements. The first author is thankful to University GrantsCommission (UGC), India for the financial support under the grant SAP-DRS-On some continued fraction expansions for the ratios of the function ρ(a, b) 411-NO.F.510/2/DRS/2011. The second author is thankful to UGC for awardingthe award of Rajiv Gandhi National Fellowship, No. F1-17.1/2011-12/RGNF-SC-KAR-2983/(SA-III/Website) and the third author is thankful to UniversityGrants Commission, India, for the award of Teacher Fellowship under the grantNo. KAMY074-TF01-13112010.References[1] C. Adiga, R. Y. Denis and K. R. Vasuki: On some continued fraction expansionsfor the ratios of 2ψ2, Special Functions: Selected Articles, P. K. Banerji (Ed.), ScientificPublishers (India), 2001, 1–16.[2] C. Adiga and D. D. Somashekara:On some Rogers-Ramanujan type continued fractionidentities, Mathematical Balcanika, New series, 12 (1998), 37–45.[3] W. A. Al-Salam and M. E. H. Ismail: Orthogonal polynomials associated with theRogers-Ramanujan continued fraction, Pacific J. Math., 104 (1983), 269–283.[4] G. E. Andrews: Ramanujan’s ‘lost’ notebook III. The Rogers-Ramanujan continued frac-tion, Adv. Math., 41 (1981), 186–208.[5] N. A. 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Effect of Mucuna pruriens seed and Glycyrrhiza glabra root extract on longevity of Drosophila melanogaster and vestigial wing mutant

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Effect of Mucuna pruriens seed and Glycyrrhiza glabra root extract on longevity of Drosophila melanogaster and vestigial wing mutant

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